# Residue theorem for evaluating integral

Calculate integral with residue theorem and make sketch of $\gamma$.

$\gamma : [0,4\pi ]\mapsto \mathbb C , t \mapsto i+2e^{it}$

$\int_{\gamma }^{} \! (e^{\frac{1}{z} }+\frac{sin(z^2)}{z+2} -\frac{5}{z-i} ) \, dz$

My try: I found singularities and residues.

$f_1(z)=e^{\frac{1}{z}}$ , $z_0=0$ , $Res=1$

$f_2(z)=\frac{sin(z^2)}{z+2}$ , $z_0=-2$ , $Res=sin(-4)$

$f_3(z)=\frac{-5}{z-i}$ , $z_0=i$ , $Res=-5$

What to do next? I know that for residue theorem I have to multiply residues with $2\pi i$. But what should I do with $\gamma$ and $t \mapsto i+2e^{it}$? How do I make sketch?

• You need to look at the mapping $t \mapsto i+2e^{it}$ and what it really represents. – Zaid Alyafeai Apr 14 '17 at 23:17

First, we notice that $\gamma$ is two complete turn arounds (as we have $4\pi$ instead of $2\pi$) in the circle trajectory around $i$. Second, the only singularities inside the circle of radius $2$ with center $i$ are $0$ and $i$, so we only have to consider the residues of these two singularities. Third and last, we make the addition of the residues with multiplicities (where these are the number of times the closed path goes around the singularity counting counter clockwise turns as positive and clockwise turns as negative); in this particular case, $\gamma$ goes around $0$ two times in a counter clockwise way and around $i$ two times in a counter clockwise way, so we get $$\frac{1}{2\pi i}\int_{\gamma }\, \left(e^{\frac{1}{z} }+\frac{\sin(z^2)}{z+2} -\frac{5}{z-i} \right) \, dz=2\mathrm{Res}_{z=0}\left(e^{\frac{1}{z} }+\frac{\sin(z^2)}{z+2} -\frac{5}{z-i} \right)+2\mathrm{Res}_{z=i}\left(e^{\frac{1}{z} }+\frac{\sin(z^2)}{z+2} -\frac{5}{z-i} \right)=2\cdot 1+2\cdot (-5)=-8\text{.}$$ Hence $$\int_{\gamma }\, \left(e^{\frac{1}{z} }+\frac{\sin(z^2)}{z+2} -\frac{5}{z-i} \right) \, dz=-16\pi i\text{.}$$ Also note, although not important for the integral as its singularity is not surrounded by $\gamma$, that $$\mathrm{Res}_{z=-2}\left(e^{\frac{1}{z} }+\frac{\sin(z^2)}{z+2} -\frac{5}{z-i} \right)=\mathrm{Res}_{z=-2}\frac{\sin(z^2)}{z+2}=\sin((-2)^2)=\sin(4)$$ and not $\sin(-4)$.